Coefficient of x^3

  1. Find the coefficient of x^3 in the expansion of (2x^2-3/x)^3


    I know how to do simple coefficients using pascalles triangle but I really don't know how to do this.
    Any help would be much appreciated.
     
  2. jcsd
  3. Write it as

    [tex](2x^2 - 3x^{-1})^3[/tex]

    From Pascal's triangle, you know how to expand

    [tex](a+b)^n[/tex]

    What can you replace with a and what can you replace with b?
     
  4. rock.freak667

    rock.freak667 6,231
    Homework Helper

    [tex](2x^2-\frac{3}{x})^3[/tex]

    [tex] (\frac{1}{x}(2x^3-3))^3[/tex]

    How about now?
     
  5. Oh, that's a nice way of doing it :)
     
  6. rock.freak667

    rock.freak667 6,231
    Homework Helper

    Usually (well for me), a binomial expansion is usually done with a variable and a constant.

    as for [itex](a+b)^n[/itex] is valid for [itex]|\frac{b}{a}|<1[/itex] But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.
     
  7. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    Why is it only valid in some range??? I also don't see why you need to factor the original. (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3. Just put a=2x^2 and b=(-3/x), figure out which term is the x^3 term and evaluate it.
     
  8. rock.freak667

    rock.freak667 6,231
    Homework Helper

    That's what I was taught.."validity of a binomial"
     
  9. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    Got a reference? If you are thinking of the convergence of the infinite series for negative exponents, that is something to think about. But this is a positive exponent, the series is finite. There are no convergence issues.
     
  10. Besides, we're dealing with polynomials in the case of (a+b)^n
     
  11. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    Yeah, that's what I mean by finite series. You could also just forget about pascal's triangle and multiply it out. The power is only 3.
     
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